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(A) Given,$v = \alpha \sqrt{x}$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \alpha \sqrt{x}$.Separating variables: $\frac{dx}{\sqrt{x}} = \alph

Vedclass · Accepted Answer

$\frac{\alpha^2}{2} t$ and $\frac{\alpha^2}{2}$

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(A) Given,$v = \alpha \sqrt{x}$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \alpha \sqrt{x}$.Separating variables: $\frac{dx}{\sqrt{x}} = \alpha dt$.Integrating both sides: $\int x^{-1/2} dx = \int \alpha dt \Rightarrow 2\sqrt{x} = \alpha t + C$.At $t = 0$,$x = 0$,so $C = 0$. Thus,$2\sqrt{x} = \alpha t \Rightarrow \sqrt{x} = \frac{\alpha t}{2}$.Squaring both sides: $x = \frac{\alpha^2 t^2}{4}$.Velocity $v = \frac{dx}{dt} = \frac{d}{dt} \left( \frac{\alpha^2 t^2}{4} \right) = \frac{\alpha^2}{4} (2t) = \frac{\alpha^2 t}{2}$.Acceleration $a = \frac{dv}{dt} = \frac{d}{dt} \left( \frac{\alpha^2 t}{2} \right) = \frac{\alpha^2}{2}$.

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